Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Metoda Statistika |
| Periode Ujian | : | November 2014 |
| Nomor Soal | : | 8 |
SOAL
Berdasarkan soal nomor 7. Tentukan \(Var\left( X \right)\)
- \(\frac{1}{2}r\)
- \(\frac{1}{4}r\)
- \(r\)
- \(2r\)
- \(\frac{1}{{r – 2}}\)
| Diketahui | \(f\left( x \right) = \frac{1}{{{2^{\frac{r}{2}}}\Gamma \left( {\frac{r}{2}} \right)}}{x^{\frac{r}{2} – 1}}{e^{ – \frac{x}{2}}},x > 0\) adalah distribusi Chi-Square dengan r degrees of freedom \(\left( {r > 0} \right)\) dan \(E\left[ X \right] = r\) |
| Rumus yang digunakan | Distribusi Gamma: \(f\left( x \right) = \frac{1}{{{\beta ^\alpha }\Gamma \left( \alpha \right)}}{x^{\alpha – 1}}{e^{ – \frac{x}{\beta }}}\) dan \(\int\limits_0^\infty {\frac{1}{{{\beta ^\alpha }\Gamma \left( \alpha \right)}}{x^{\alpha – 1}}{e^{ – \frac{x}{\beta }}}dx} = 1 \Leftrightarrow \int\limits_0^\infty {{x^{\alpha – 1}}{e^{ – \frac{x}{\beta }}}dx} = {\beta ^\alpha }\Gamma \left( \alpha \right)\)
Fungsi Gamma: \(\Gamma \left( \alpha \right) = \int\limits_0^\infty {{x^{\alpha – 1}}{e^{ – x}}dx} \) atau \(\Gamma \left( \alpha \right) = \left( {\alpha – 1} \right)!\) \(E\left[ X \right] = \int\limits_{ – \infty }^\infty {xf\left( x \right)dx} \) |
| Proses pengerjaan | Kita misalkan \(\alpha = \frac{r}{2}\) dan \(\beta = 2\) sehingga menjadi distribusi gamma, maka \(E\left[ {{X^2}} \right] = \int\limits_{ – \infty }^\infty {{x^2}f\left( x \right)dx} = \int\limits_0^\infty {{x^2} \cdot \frac{1}{{{\beta ^\alpha }\Gamma \left( \alpha \right)}}{x^{\alpha – 1}}{e^{ – \frac{x}{\beta }}}dx} \) \(E\left[ {{X^2}} \right] = \int\limits_{ – \infty }^\infty {{x^2}f\left( x \right)dx} = \int\limits_0^\infty {{x^2} \cdot \frac{1}{{{\beta ^\alpha }\Gamma \left( \alpha \right)}}{x^{\alpha – 1}}{e^{ – \frac{x}{\beta }}}dx} \) \(E\left[ {{X^2}} \right] = \frac{1}{{{\beta ^\alpha }\Gamma \left( \alpha \right)}}\int\limits_0^\infty {{x^{\alpha + 1}}{e^{ – \frac{x}{\beta }}}dx} \) \(E\left[ {{X^2}} \right] = \frac{1}{{{\beta ^\alpha }\Gamma \left( \alpha \right)}}\left[ {{\beta ^{\alpha + 2}}\Gamma \left( {\alpha + 2} \right)} \right]\) \(E\left[ {{X^2}} \right] = \frac{{{\beta ^2}\left[ {\left( {\alpha + 1} \right)!} \right]}}{{\left( {\alpha – 1} \right)!}} = \frac{{{\beta ^2}\left( {\alpha + 1} \right)\alpha \left[ {\left( {\alpha – 1} \right)!} \right]}}{{\left( {\alpha – 1} \right)!}}\) \(E\left[ {{X^2}} \right] = {\beta ^2}\left( {\alpha + 1} \right)\alpha = {2^2} \cdot \left( {\frac{r}{2} + 1} \right) \cdot \left( {\frac{r}{2}} \right) = {r^2} + 2r\) \(Var\left( X \right) = E\left[ {{X^2}} \right] – {\left( {E\left[ X \right]} \right)^2} = \left( {{r^2} + 2r} \right) – {r^2} = 2r\) |
| Jawaban | D. \(2r\) |
